Advanced Calculus Single Variable

7.11 Derivatives And Limits Of Sequences

When you have a function which is a limit of a sequence of functions, when can you say the
derivative of the limit function is the limit of the derivatives of the functions in the sequence?
The following theorem seems to be one of the best results available. It is based on the
mean value theorem. First of all, recall Definition 6.9.6 on Page 298 listed here for
convenience.

Definition 7.11.1Let

{fn}

be a sequence of functions defined on D. Then

{fn}

issaid to converge uniformly to f if it convergespointwise to f and for every ε > 0 there existsN such that for all n ≥ N

|f (x)− fn(x)| < ε

for all x ∈ D.

To save on notation, denote by

||k|| ≡ sup{|k(ξ)| : ξ ∈ D }.

Then

||k + l|| ≤ ||k||+||l|| (7.18)

(7.18)

because for each ξ ∈ D,

|k(ξ)+ l(ξ)| ≤ ||k||+ ||l||

and taking sup yields 7.18. From the definition of uniform convergence, you see that fn
converges uniformly to f is the same as saying

lim ||fn − f|| = 0.
n→∞

Now here is the theorem. Note how the mean value theorem is one of the principal parts of
the argument.

Theorem 7.11.2Let

(a,b)

be a finiteopen interval and let fk :

(a,b)

→ ℝ bedifferentiable and suppose there exists x0∈

(a,b)

such that

{fk (x0)} converges,

{fk′} converges uniformly to a function g on (a,b).

Then there exists a function f defined on

(a,b)

such that

fk → f uniformly,

and

f′ = g.

Proof: Let c ∈

(a,b)

and define

{ fn(x)−fn(c)if x ⁄= c
gn (x,c) ≡ f′(xc−)c if x = c .
n

Also let

||h|| ≡ sup {|h (x)| : x ∈ (a,b)}.

Thus hk→ h uniformly means

||hk − h||

→ 0.

Claim 1: For each c, x → gn

(x,c)

converges uniformly to a continuous function hc, on

(a,b)

and hc

(c)

= g

(c)

.

Proof: First note that each x → gn

(x,c)

is continuous. Next consider the claim about
uniform convergence. Let x≠c. Then by the mean value theorem applied to the function
x → fn